{"id":207,"date":"2025-02-13T22:45:03","date_gmt":"2025-02-13T22:45:03","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/modeling-with-trigonometric-equations\/"},"modified":"2025-10-16T19:02:08","modified_gmt":"2025-10-16T19:02:08","slug":"modeling-with-trigonometric-equations","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/modeling-with-trigonometric-equations\/","title":{"raw":"Modeling with Trigonometric Equations: Learn It 1","rendered":"Modeling with Trigonometric Equations: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li style=\"font-weight: 400;\">Determine the amplitude and period of a periodic context<\/li>\r\n \t<li style=\"font-weight: 400;\">Model periodic behavior with sinusoidal functions<\/li>\r\n \t<li style=\"font-weight: 400;\">Write both a sine and cosine function to model the same periodic behavior<\/li>\r\n<\/ul>\r\n<\/section>Any motion that repeats itself in a fixed time period is considered <strong>periodic motion<\/strong> and can be modeled by a <strong>sinusoidal function<\/strong>. \u00a0Sinusoidal functions oscillate above and below the midline, are periodic, and repeat values in set cycles.\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">The <strong>amplitude<\/strong> of a sinusoidal function is the distance from the midline to the maximum value, or from the midline to the minimum value. The <strong>midline<\/strong> is the average value.<\/section><section class=\"textbox recall\" aria-label=\"Recall\">The <strong>period<\/strong> of the sine function and the cosine function is [latex]\\text{ }2\\pi .\\text{ }[\/latex] In other words, for any value of [latex]\\text{ }x[\/latex],\r\n<div style=\"text-align: center;\">[latex]\\sin \\left(x\\pm 2\\pi k\\right)=\\sin x\\text{ and }\\cos \\left(x\\pm 2\\pi k\\right)=\\cos x\\text{ where }k\\text{ is an integer}[\/latex]<\/div>\r\n<\/section>\r\n<div><section class=\"textbox recall\" aria-label=\"Recall\">The general forms of a sinusoidal equation are given as\r\n<div style=\"text-align: center;\">[latex]y=A\\sin \\left(Bt-C\\right)+D\\text{ or }y=A\\cos \\left(Bt-C\\right)+D[\/latex]<\/div>\r\nwhere [latex]\\text{amplitude}=|A|,B[\/latex] is related to period such that the [latex]\\text{ period}=\\frac{2\\pi }{B},C\\text{ }[\/latex] is the phase shift such that [latex]\\text{ }\\frac{C}{B}\\text{ }[\/latex] denotes the horizontal shift, and [latex]\\text{ }D\\text{ }[\/latex] represents the vertical shift from the graph\u2019s parent graph.\r\n\r\nThe difference between the sine and the cosine graphs is that the sine graph begins with the average value of the function and the cosine graph begins with the maximum or minimum value of the function.\r\n\r\n<\/section>\r\n<h2>Modeling Periodic Behavior<\/h2>\r\n<section class=\"textbox example\" aria-label=\"Example\">The average monthly temperatures for a small town in Oregon are given in the table below. Find a sinusoidal function of the form [latex]y=A\\sin \\left(Bt-C\\right)+D[\/latex] that fits the data (round to the nearest tenth) and sketch the graph.\r\n<table id=\"fs-id1336940\" summary=\"Thirteen rows, two columns. The table has ordered pairs of these row values: (Month, Temperature in degrees F), (January, 42.5), (February, 44.5), (March, 48.5), (April, 52.5), (May, 58), (June, 63), (July, 68.5), (August, 69), (September, 64.5), (October, 55.5), (November, 46.5), (December, 43.5).\"><colgroup> <col \/> <col \/> <\/colgroup>\r\n<thead>\r\n<tr>\r\n<th>Month<\/th>\r\n<th>Temperature, [latex]{}^{\\text{o}}\\text{F}[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>January<\/td>\r\n<td>42.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>February<\/td>\r\n<td>44.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>March<\/td>\r\n<td>48.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>April<\/td>\r\n<td>52.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>May<\/td>\r\n<td>58<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>June<\/td>\r\n<td>63<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>July<\/td>\r\n<td>68.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>August<\/td>\r\n<td>69<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>September<\/td>\r\n<td>64.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>October<\/td>\r\n<td>55.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>November<\/td>\r\n<td>46.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>December<\/td>\r\n<td>43.5<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"270264\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"270264\"]\r\n\r\nRecall that amplitude is found using the formula\r\n<p style=\"text-align: center;\">[latex]A=\\frac{\\text{largest value }-\\text{smallest value}}{2}[\/latex]<\/p>\r\nThus, the amplitude is\r\n<p style=\"text-align: center;\">[latex]\\begin{align} |A|&amp;=\\frac{69 - 42.5}{2} \\\\ &amp;=13.25 \\end{align}[\/latex]<\/p>\r\nThe data covers a period of 12 months, so [latex]\\frac{2\\pi }{B}=12[\/latex] which gives [latex]B=\\frac{2\\pi }{12}=\\frac{\\pi }{6}[\/latex].\r\n\r\nThe vertical shift is found using the following equation.\r\n<p style=\"text-align: center;\">[latex]D=\\frac{\\text{highest value}+\\text{lowest value}}{2}[\/latex]<\/p>\r\nThus, the vertical shift is\r\n<p style=\"text-align: center;\">[latex]\\begin{align}D&amp;=\\frac{69+42.5}{2} \\\\ &amp;=55.8\\end{align}[\/latex]<\/p>\r\nSo far, we have the equation [latex]y=13.3\\sin \\left(\\frac{\\pi }{6}x-C\\right)+55.8[\/latex].\r\n\r\nTo find the horizontal shift, we input the [latex]x[\/latex] and [latex]y[\/latex] values for the first month and solve for [latex]C[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}42.5=13.3\\sin \\left(\\frac{\\pi }{6}\\left(1\\right)-C\\right)+55.8 \\\\ -13.3=13.3\\sin \\left(\\frac{\\pi }{6}-C\\right) \\\\ -1=\\sin \\left(\\frac{\\pi }{6}-C\\right) \\\\ \\sin \\theta =-1\\to \\theta =-\\frac{\\pi }{2}\\end{gathered}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align} \\frac{\\pi }{6}-C&amp;=-\\frac{\\pi }{2}\\\\ \\frac{\\pi }{6}+\\frac{\\pi }{2}&amp;=C \\\\ &amp;=\\frac{2\\pi }{3} \\end{align}[\/latex]<\/p>\r\nWe have the equation [latex]y=13.3\\sin \\left(\\frac{\\pi }{6}x-\\frac{2\\pi }{3}\\right)+55.8[\/latex]. See the graph in Figure 8.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164135\/CNX_Precalc_Figure_07_06_0112.jpg\" alt=\"Graph of the equation y=13.3sin(pi\/6 x - 2pi\/3) + 55.8. The average value is a dotted horizontal line y=55.8, and the amplitude is 13.3\" width=\"487\" height=\"331\" \/> <b>Figure 8<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">The hour hand of the large clock on the wall in Union Station measures 24 inches in length. At noon, the tip of the hour hand is 30 inches from the ceiling. At 3 PM, the tip is 54 inches from the ceiling, and at 6 PM, 78 inches. At 9 PM, it is again 54 inches from the ceiling, and at midnight, the tip of the hour hand returns to its original position 30 inches from the ceiling. Let [latex]y[\/latex] equal the distance from the tip of the hour hand to the ceiling [latex]x[\/latex] hours after noon. Find the equation that models the motion of the clock and sketch the graph.[reveal-answer q=\"45024\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"45024\"]Begin by making a table of values as shown in the table below.\r\n<table id=\"fs-id1992294\" summary=\"Six rows, three columns. The table has ordered pairs of these row values: (X, y, Points to plot), (Noon, 30 in, (0,30)), (3 PM, 54 in, (3,54)), (6 PM, 78 in, (6,8)), (9 PM, 54 in, (9,54)), (Midnight, 30 in, (12,30)).\"><colgroup> <col \/> <col \/> <col \/> <\/colgroup>\r\n<thead>\r\n<tr>\r\n<th>[latex]x[\/latex]<\/th>\r\n<th>[latex]y[\/latex]<\/th>\r\n<th>Points to plot<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Noon<\/td>\r\n<td>30 in<\/td>\r\n<td>[latex]\\left(0,30\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3 PM<\/td>\r\n<td>54 in<\/td>\r\n<td>[latex]\\left(3,54\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>6 PM<\/td>\r\n<td>78 in<\/td>\r\n<td>[latex]\\left(6,78\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>9 PM<\/td>\r\n<td>54 in<\/td>\r\n<td>[latex]\\left(9,54\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Midnight<\/td>\r\n<td>30 in<\/td>\r\n<td>[latex]\\left(12,30\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nTo model an equation, we first need to find the amplitude.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}|A|&amp;=\\left\\rvert\\frac{78 - 30}{2}\\right\\rvert \\\\ &amp;=24 \\end{align}[\/latex]<\/p>\r\nThe clock\u2019s cycle repeats every 12 hours. Thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{align}B&amp;=\\frac{2\\pi }{12} \\\\ &amp;=\\frac{\\pi }{6} \\end{align}[\/latex]<\/p>\r\nThe vertical shift is\r\n<p style=\"text-align: center;\">[latex]\\begin{align}D&amp;=\\frac{78+30}{2} \\\\ &amp;=54 \\end{align}[\/latex]<\/p>\r\nThere is no horizontal shift, so [latex]C=0[\/latex]. Since the function begins with the minimum value of [latex]y[\/latex] when [latex]x=0[\/latex] (as opposed to the maximum value), we will use the cosine function with the negative value for [latex]A[\/latex]. In the form [latex]y=A\\cos \\left(Bx\\pm C\\right)+D[\/latex], the equation is\r\n<p style=\"text-align: center;\">[latex]y=-24\\cos \\left(\\frac{\\pi }{6}x\\right)+54[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164137\/CNX_Precalc_Figure_07_06_0132.jpg\" alt=\"Graph of the function y=-24cos(pi\/6 x)+54 using the five key points: (0,30), (3,54), (6,78), (9,54), (12,30).\" width=\"487\" height=\"392\" \/> <b>Figure 9<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">The height of the tide in a small beach town is measured along a seawall. Water levels oscillate between 7 feet at low tide and 15 feet at high tide. On a particular day, low tide occurred at 6 AM and high tide occurred at noon. Approximately every 12 hours, the cycle repeats. Find an equation to model the water levels.[reveal-answer q=\"705973\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"705973\"]As the water level varies from 7 ft to 15 ft, we can calculate the amplitude as\r\n<p style=\"text-align: center;\">[latex]\\begin{align}|A|&amp;=\\left\\rvert\\frac{\\left(15 - 7\\right)}{2}\\right\\rvert \\\\ &amp;=4 \\end{align}[\/latex]<\/p>\r\nThe cycle repeats every 12 hours; therefore, [latex]B[\/latex] is\r\n<p style=\"text-align: center;\">[latex]\\frac{2\\pi }{12}=\\frac{\\pi }{6}[\/latex]<\/p>\r\nThere is a vertical translation of [latex]\\frac{\\left(15+8\\right)}{2}=11.5[\/latex]. Since the value of the function is at a maximum at [latex]t=0[\/latex], we will use the cosine function, with the positive value for [latex]A[\/latex].\r\n<p style=\"text-align: center;\">[latex]y=4\\cos \\left(\\frac{\\pi }{6}\\right)t+11[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164139\/CNX_Precalc_Figure_07_06_0152.jpg\" alt=\"Graph of the function y=4cos(pi\/6 t) + 11 from 0 to 12. The midline is y=11, three key points are (0,15), (6,7), and (12, 15).\" width=\"487\" height=\"330\" \/> <b>Figure 10<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<div class=\"bcc-box bcc-success\">\r\n\r\nThe daily temperature in the month of March in a certain city varies from a low of [latex]24^\\circ\\text{F}[\/latex] to a high of [latex]40^\\circ\\text{F}[\/latex]. Find a sinusoidal function to model daily temperature and sketch the graph. Approximate the time when the temperature reaches the freezing point [latex]32^\\circ\\text{F}[\/latex]. Let [latex]t=0[\/latex] correspond to noon.\r\n\r\n[reveal-answer q=\"130521\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"130521\"]\r\n\r\n[latex]y=8\\sin \\left(\\frac{\\pi }{12}t\\right)+32[\/latex]\r\nThe temperature reaches freezing at noon and at midnight.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164214\/CNX_Precalc_Figure_07_06_0162.jpg\" alt=\"Graph of the function y=8sin(pi\/12 t) + 32 for temperature. The midline is at 32. The times when the temperature is at 32 are midnight and noon.\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>\r\n<div class=\"bcc-box bcc-success\"><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]152828[\/ohm_question]<\/section><\/div>\r\n<\/div>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li style=\"font-weight: 400;\">Determine the amplitude and period of a periodic context<\/li>\n<li style=\"font-weight: 400;\">Model periodic behavior with sinusoidal functions<\/li>\n<li style=\"font-weight: 400;\">Write both a sine and cosine function to model the same periodic behavior<\/li>\n<\/ul>\n<\/section>\n<p>Any motion that repeats itself in a fixed time period is considered <strong>periodic motion<\/strong> and can be modeled by a <strong>sinusoidal function<\/strong>. \u00a0Sinusoidal functions oscillate above and below the midline, are periodic, and repeat values in set cycles.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">The <strong>amplitude<\/strong> of a sinusoidal function is the distance from the midline to the maximum value, or from the midline to the minimum value. The <strong>midline<\/strong> is the average value.<\/section>\n<section class=\"textbox recall\" aria-label=\"Recall\">The <strong>period<\/strong> of the sine function and the cosine function is [latex]\\text{ }2\\pi .\\text{ }[\/latex] In other words, for any value of [latex]\\text{ }x[\/latex],<\/p>\n<div style=\"text-align: center;\">[latex]\\sin \\left(x\\pm 2\\pi k\\right)=\\sin x\\text{ and }\\cos \\left(x\\pm 2\\pi k\\right)=\\cos x\\text{ where }k\\text{ is an integer}[\/latex]<\/div>\n<\/section>\n<div>\n<section class=\"textbox recall\" aria-label=\"Recall\">The general forms of a sinusoidal equation are given as<\/p>\n<div style=\"text-align: center;\">[latex]y=A\\sin \\left(Bt-C\\right)+D\\text{ or }y=A\\cos \\left(Bt-C\\right)+D[\/latex]<\/div>\n<p>where [latex]\\text{amplitude}=|A|,B[\/latex] is related to period such that the [latex]\\text{ period}=\\frac{2\\pi }{B},C\\text{ }[\/latex] is the phase shift such that [latex]\\text{ }\\frac{C}{B}\\text{ }[\/latex] denotes the horizontal shift, and [latex]\\text{ }D\\text{ }[\/latex] represents the vertical shift from the graph\u2019s parent graph.<\/p>\n<p>The difference between the sine and the cosine graphs is that the sine graph begins with the average value of the function and the cosine graph begins with the maximum or minimum value of the function.<\/p>\n<\/section>\n<h2>Modeling Periodic Behavior<\/h2>\n<section class=\"textbox example\" aria-label=\"Example\">The average monthly temperatures for a small town in Oregon are given in the table below. Find a sinusoidal function of the form [latex]y=A\\sin \\left(Bt-C\\right)+D[\/latex] that fits the data (round to the nearest tenth) and sketch the graph.<\/p>\n<table id=\"fs-id1336940\" summary=\"Thirteen rows, two columns. The table has ordered pairs of these row values: (Month, Temperature in degrees F), (January, 42.5), (February, 44.5), (March, 48.5), (April, 52.5), (May, 58), (June, 63), (July, 68.5), (August, 69), (September, 64.5), (October, 55.5), (November, 46.5), (December, 43.5).\">\n<colgroup>\n<col \/>\n<col \/> <\/colgroup>\n<thead>\n<tr>\n<th>Month<\/th>\n<th>Temperature, [latex]{}^{\\text{o}}\\text{F}[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>January<\/td>\n<td>42.5<\/td>\n<\/tr>\n<tr>\n<td>February<\/td>\n<td>44.5<\/td>\n<\/tr>\n<tr>\n<td>March<\/td>\n<td>48.5<\/td>\n<\/tr>\n<tr>\n<td>April<\/td>\n<td>52.5<\/td>\n<\/tr>\n<tr>\n<td>May<\/td>\n<td>58<\/td>\n<\/tr>\n<tr>\n<td>June<\/td>\n<td>63<\/td>\n<\/tr>\n<tr>\n<td>July<\/td>\n<td>68.5<\/td>\n<\/tr>\n<tr>\n<td>August<\/td>\n<td>69<\/td>\n<\/tr>\n<tr>\n<td>September<\/td>\n<td>64.5<\/td>\n<\/tr>\n<tr>\n<td>October<\/td>\n<td>55.5<\/td>\n<\/tr>\n<tr>\n<td>November<\/td>\n<td>46.5<\/td>\n<\/tr>\n<tr>\n<td>December<\/td>\n<td>43.5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q270264\">Show Solution<\/button><\/p>\n<div id=\"q270264\" class=\"hidden-answer\" style=\"display: none\">\n<p>Recall that amplitude is found using the formula<\/p>\n<p style=\"text-align: center;\">[latex]A=\\frac{\\text{largest value }-\\text{smallest value}}{2}[\/latex]<\/p>\n<p>Thus, the amplitude is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} |A|&=\\frac{69 - 42.5}{2} \\\\ &=13.25 \\end{align}[\/latex]<\/p>\n<p>The data covers a period of 12 months, so [latex]\\frac{2\\pi }{B}=12[\/latex] which gives [latex]B=\\frac{2\\pi }{12}=\\frac{\\pi }{6}[\/latex].<\/p>\n<p>The vertical shift is found using the following equation.<\/p>\n<p style=\"text-align: center;\">[latex]D=\\frac{\\text{highest value}+\\text{lowest value}}{2}[\/latex]<\/p>\n<p>Thus, the vertical shift is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}D&=\\frac{69+42.5}{2} \\\\ &=55.8\\end{align}[\/latex]<\/p>\n<p>So far, we have the equation [latex]y=13.3\\sin \\left(\\frac{\\pi }{6}x-C\\right)+55.8[\/latex].<\/p>\n<p>To find the horizontal shift, we input the [latex]x[\/latex] and [latex]y[\/latex] values for the first month and solve for [latex]C[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}42.5=13.3\\sin \\left(\\frac{\\pi }{6}\\left(1\\right)-C\\right)+55.8 \\\\ -13.3=13.3\\sin \\left(\\frac{\\pi }{6}-C\\right) \\\\ -1=\\sin \\left(\\frac{\\pi }{6}-C\\right) \\\\ \\sin \\theta =-1\\to \\theta =-\\frac{\\pi }{2}\\end{gathered}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} \\frac{\\pi }{6}-C&=-\\frac{\\pi }{2}\\\\ \\frac{\\pi }{6}+\\frac{\\pi }{2}&=C \\\\ &=\\frac{2\\pi }{3} \\end{align}[\/latex]<\/p>\n<p>We have the equation [latex]y=13.3\\sin \\left(\\frac{\\pi }{6}x-\\frac{2\\pi }{3}\\right)+55.8[\/latex]. See the graph in Figure 8.<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164135\/CNX_Precalc_Figure_07_06_0112.jpg\" alt=\"Graph of the equation y=13.3sin(pi\/6 x - 2pi\/3) + 55.8. The average value is a dotted horizontal line y=55.8, and the amplitude is 13.3\" width=\"487\" height=\"331\" \/><figcaption class=\"wp-caption-text\"><b>Figure 8<\/b><\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">The hour hand of the large clock on the wall in Union Station measures 24 inches in length. At noon, the tip of the hour hand is 30 inches from the ceiling. At 3 PM, the tip is 54 inches from the ceiling, and at 6 PM, 78 inches. At 9 PM, it is again 54 inches from the ceiling, and at midnight, the tip of the hour hand returns to its original position 30 inches from the ceiling. Let [latex]y[\/latex] equal the distance from the tip of the hour hand to the ceiling [latex]x[\/latex] hours after noon. Find the equation that models the motion of the clock and sketch the graph.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q45024\">Show Solution<\/button><\/p>\n<div id=\"q45024\" class=\"hidden-answer\" style=\"display: none\">Begin by making a table of values as shown in the table below.<\/p>\n<table id=\"fs-id1992294\" summary=\"Six rows, three columns. The table has ordered pairs of these row values: (X, y, Points to plot), (Noon, 30 in, (0,30)), (3 PM, 54 in, (3,54)), (6 PM, 78 in, (6,8)), (9 PM, 54 in, (9,54)), (Midnight, 30 in, (12,30)).\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/> <\/colgroup>\n<thead>\n<tr>\n<th>[latex]x[\/latex]<\/th>\n<th>[latex]y[\/latex]<\/th>\n<th>Points to plot<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Noon<\/td>\n<td>30 in<\/td>\n<td>[latex]\\left(0,30\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>3 PM<\/td>\n<td>54 in<\/td>\n<td>[latex]\\left(3,54\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>6 PM<\/td>\n<td>78 in<\/td>\n<td>[latex]\\left(6,78\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>9 PM<\/td>\n<td>54 in<\/td>\n<td>[latex]\\left(9,54\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Midnight<\/td>\n<td>30 in<\/td>\n<td>[latex]\\left(12,30\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>To model an equation, we first need to find the amplitude.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}|A|&=\\left\\rvert\\frac{78 - 30}{2}\\right\\rvert \\\\ &=24 \\end{align}[\/latex]<\/p>\n<p>The clock\u2019s cycle repeats every 12 hours. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}B&=\\frac{2\\pi }{12} \\\\ &=\\frac{\\pi }{6} \\end{align}[\/latex]<\/p>\n<p>The vertical shift is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}D&=\\frac{78+30}{2} \\\\ &=54 \\end{align}[\/latex]<\/p>\n<p>There is no horizontal shift, so [latex]C=0[\/latex]. Since the function begins with the minimum value of [latex]y[\/latex] when [latex]x=0[\/latex] (as opposed to the maximum value), we will use the cosine function with the negative value for [latex]A[\/latex]. In the form [latex]y=A\\cos \\left(Bx\\pm C\\right)+D[\/latex], the equation is<\/p>\n<p style=\"text-align: center;\">[latex]y=-24\\cos \\left(\\frac{\\pi }{6}x\\right)+54[\/latex]<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164137\/CNX_Precalc_Figure_07_06_0132.jpg\" alt=\"Graph of the function y=-24cos(pi\/6 x)+54 using the five key points: (0,30), (3,54), (6,78), (9,54), (12,30).\" width=\"487\" height=\"392\" \/><figcaption class=\"wp-caption-text\"><b>Figure 9<\/b><\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">The height of the tide in a small beach town is measured along a seawall. Water levels oscillate between 7 feet at low tide and 15 feet at high tide. On a particular day, low tide occurred at 6 AM and high tide occurred at noon. Approximately every 12 hours, the cycle repeats. Find an equation to model the water levels.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q705973\">Show Solution<\/button><\/p>\n<div id=\"q705973\" class=\"hidden-answer\" style=\"display: none\">As the water level varies from 7 ft to 15 ft, we can calculate the amplitude as<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}|A|&=\\left\\rvert\\frac{\\left(15 - 7\\right)}{2}\\right\\rvert \\\\ &=4 \\end{align}[\/latex]<\/p>\n<p>The cycle repeats every 12 hours; therefore, [latex]B[\/latex] is<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{2\\pi }{12}=\\frac{\\pi }{6}[\/latex]<\/p>\n<p>There is a vertical translation of [latex]\\frac{\\left(15+8\\right)}{2}=11.5[\/latex]. Since the value of the function is at a maximum at [latex]t=0[\/latex], we will use the cosine function, with the positive value for [latex]A[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]y=4\\cos \\left(\\frac{\\pi }{6}\\right)t+11[\/latex]<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164139\/CNX_Precalc_Figure_07_06_0152.jpg\" alt=\"Graph of the function y=4cos(pi\/6 t) + 11 from 0 to 12. The midline is y=11, three key points are (0,15), (6,7), and (12, 15).\" width=\"487\" height=\"330\" \/><figcaption class=\"wp-caption-text\"><b>Figure 10<\/b><\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<div class=\"bcc-box bcc-success\">\n<p>The daily temperature in the month of March in a certain city varies from a low of [latex]24^\\circ\\text{F}[\/latex] to a high of [latex]40^\\circ\\text{F}[\/latex]. Find a sinusoidal function to model daily temperature and sketch the graph. Approximate the time when the temperature reaches the freezing point [latex]32^\\circ\\text{F}[\/latex]. Let [latex]t=0[\/latex] correspond to noon.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q130521\">Show Solution<\/button><\/p>\n<div id=\"q130521\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y=8\\sin \\left(\\frac{\\pi }{12}t\\right)+32[\/latex]<br \/>\nThe temperature reaches freezing at noon and at midnight.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164214\/CNX_Precalc_Figure_07_06_0162.jpg\" alt=\"Graph of the function y=8sin(pi\/12 t) + 32 for temperature. The midline is at 32. The times when the temperature is at 32 are midnight and noon.\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"bcc-box bcc-success\">\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm152828\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=152828&theme=lumen&iframe_resize_id=ohm152828&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/div>\n<\/div>\n","protected":false},"author":6,"menu_order":33,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":201,"module-header":"learn_it","content_attributions":[{"type":"cc-attribution","description":"Precalculus","author":"OpenStax College","organization":"OpenStax","url":"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface","project":"","license":"cc-by","license_terms":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/207"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":11,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/207\/revisions"}],"predecessor-version":[{"id":4697,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/207\/revisions\/4697"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/201"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/207\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=207"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=207"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=207"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=207"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}